1. Fig. 15-1 The location of a particular gene can be seen by tagging isolated chromosomes with a fluorescent dye that highlights the gene

2. Fig. 15-4a EXPERIMENT P Generation F1F1 All offspring had red eyes 

3. Fig. 15-4b RESULTS Generation F2F2

4. Fig. 15-4c Eggs F1F1 CONCLUSION Generation P X X w Sperm X Y + + + + + Eggs Sperm + + + + + Generation F2F2  w w w w w w w w w w w w w w w

5. Fig. 15-7 (a)(b) (c) XNXNXNXN XnYXnY XNXnXNXn   XNYXNYXNXnXNXn  XnYXnY YXnXn Sperm Y XNXN Y XnXn XNXnXNXn Eggs XNXN XNXN XNXnXNXn XNYXNY XNYXNY XNXN XnXn XNXNXNXN XnXNXnXN XNYXNY XnYXnY XNXN XnXn XNXnXNXn XnXnXnXn XNYXNY XnYXnY The transmission of sex linked recessive genes

6. Sex-linked genes follow specific patterns of inheritance For a recessive sex-linked trait to be expressed – A female needs two copies of the allele – A male needs only one copy of the allele Sex-linked recessive disorders are much more common in males than in females Copyright © 2008 Pearson Education Inc., publishing as Pearson Benjamin Cummings

7. Partial Linkage Linkage is different from sex linkage Linked genes tend to be inherited together because they are located near each other on the same chromosome. Results from genes being closely linked on the same chromosome Linked genes in genetic experiments deviate from the results expected from Mendel’s law of independent assortment.

8. Dihybrid Testcross to Detect Independent Assort YYRR X yyrr YrRr X yyrr YRYryRyr yyRr Yyrr yyrr YyRr Eggs Sperm yr YR yr Yr yR Phenotypic ratio 1:1:1:1 Ratio of parental:Recombinant 1:1 Dihybrid

9. Fig. 15-9-1 EXPERIMENT P Generation (homozygous) Wild type (gray body, normal wings ) Double mutant (black body, vestigial wings)  b b vg vg b + b + vg + vg + Morgan 1912

10. Fig. 15-9-2 EXPERIMENT P Generation (homozygous) Wild type (gray body, normal wings ) Double mutant (black body, vestigial wings)  b b vg vg Double mutant TESTCROSS  b + b + vg + vg + F 1 dihybrid (wild type) b + b vg + vg

11. Fig. 15-9-3 EXPERIMENT P Generation (homozygous) Wild type (gray body, normal wings ) Double mutant (black body, vestigial wings)  b b vg vg Double mutant TESTCROSS  b + b + vg + vg + F 1 dihybrid (wild type) b + b vg + vg Testcross offspring Eggs b + vg + b vg b + vg b vg + Black- normal Gray- vestigial Black- vestigial Wild type (gray-normal) b vg Sperm b + b vg + vg b b vg vg b + b vg vg b b vg + vg

12. Fig. 15-9-4 EXPERIMENT P Generation (homozygous) RESULTS Wild type (gray body, normal wings ) Double mutant (black body, vestigial wings)  b b vg vg Double mutant TESTCROSS  b + b + vg + vg + F 1 dihybrid (wild type) b + b vg + vg Testcross offspring Eggs b + vg + b vg b + vg b vg + Black- normal Gray- vestigial Black- vestigial Wild type (gray-normal) b vg Sperm b + b vg + vg b b vg vg b + b vg vg b b vg + vg PREDICTED RATIOS If genes are located on different chromosomes: If genes are located on the same chromosome and parental alleles are always inherited together: 1 1 1 1 1 1 0 0 965 944206 185 : : : : : : : : :

13. Fig. 15-10 Testcross parents Replication of chromo- somes Gray body, normal wings (F 1 dihybrid) Black body, vestigial wings (double mutant) Replication of chromo- somes b + vg + b vg b + vg + b + vg b vg + b vg Recombinant chromosomes Meiosis I and II Meiosis I Meiosis II b vg + b + vg b vg b + vg + Eggs Testcross offspring 965 Wild type (gray-normal) 944 Black- vestigial 206 Gray- vestigial 185 Black- normal b + vg + b vg b + vg b vg b vg + Sperm b vg Parental-type offspringRecombinant offspring Recombination frequency = 391 recombinants 2,300 total offspring  100 = 17%

14. 20% recombination A B a b With crossing over A B a b Parental a B A b Recombinants 40% 10%

15. Testing for Assortment/Linkage 1.Generate a dihybrid 2.Testcross the dihybrid 3.Compare the % of parental to recombinants A.If 50% parental:50% recombinant – Independent Assortment B.If more parental than recombinant – partial linkage C.If only parental and no recombinant – complete linkage

16. The discovery of linked genes and recombination due to crossing over led Alfred Strutevant to a method of constructing genetic maps He assumed the farther apart genes are, the higher the probability that a cross over will happen between them and therefore the higher the recombination frequency.

17. The closer the two genes are on a chromosome the fewer recombinants Minimum = 0% recombinants The further two genes are on a chromosome the more recombinants Maximum = 50% recombinants Linkage therefore can be used as a measure of genetic distance on chromosome 1 Map Unit = 1 % recombination

18. Gene Order on Chromosome B – Vg 17 MU B – Cn 9 MU Vg – Cn 9.5 MU

20. Partial Linkage – two genes are so close on the same chromosome that recombination occurs less than 50% of the time. Complete Linkage – two genes on the same chromosome so close that recombination cannot separate them. Independent Assortment – two genes on different chromosomes or two genes on the same chromosome but far enough apart that recombinant occurs 50% of the time.

21. Example Problem In Drosophila long wings is dominant to dumpy wings and round eyes is dominant to star eyes. A dihybrid fly was generated by mating a long wing round eye fly with a dumpy wing star eye fly. This dihybrid fly was testcrossed and the following progeny were generated. 222 long wing round eye 215 dumpy wing star eye 33 long wing star eye 30 dumpy wing round eye a.Are these genes completely linked or partially linked? b.What is the genetic distance between these two genes? c.How would the results have differed if the genes independently assorted?

22. Exception to chromosomal Inheritance (Organellar Genes) The inheritance of traits controlled by genes present in the chloroplasts or mitochondria – Depends solely on the maternal parent because the zygote’s cytoplasm comes from the egg – Maternal Inheritance

23. Pedigree Symbols

24. Nuclear vs Organellar

25. Human Genetics Pedigree Analysis

26. Autosomal vs Sex Linked

27. Multifactorial Traits Heart disease Personality IQ

28. Alterations of chromosome number or structure cause some genetic disorders. So far we’ve seen that the phenotype can be affected by small scale changes involving individual genes Random mutations are the source of all new alleles, which can lead to a new phenotype.

29. Abnormal chromosome #: Aneuploidy

30. Human Aneuploids Trisomy 21 Sex chromosome – XO – turner syndrome – XXY – klinefelters – XYY

31. Abnormal chromosome numbers Polyploidy: Common in plant ~70 % of flowering plants, Banana are triploid, Wheat 6n Strawberries 8n

32. Alterations in chromosome structure Meiosis errors and damaging agents such as radiation can cause breakage of the chromosome four types of structural damage

33. Chromosome Structure reciprocal translocation between 9 and 22 (Philadelphia Chromosome)

34. Disorders caused by structurally altered chromosomes Cri du chat – deletion in chromosome 5 Chronic myogenous leukemia Normal chromosome 9 Normal chromosome 22 Reciprocal translocation Translocated chromosome 9 Translocated chromosome 22

35. Fig. 15-8 X chromosomes Early embryo: Allele for orange fur Allele for black fur Cell division and X chromosome inactivation Two cell populations in adult cat: Active X Inactive X Black furOrange fur Barr Body – inactive X visible in interphase nucleus

36. Genomic imprinting Def: a parental effect on gene expression Identical alleles may have different effects on offspring, depending on whether they arrive in the zygote via the ovum or via the sperm. Fragile X syndrome: higher prevalence of disorder and retardation in males